The lie-superalgebras tag has no usage guidance.

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### Why, in terms of quantum groups, does the knot determinant appear as an evaluation of both the Jones and Alexander polynomials?

The Jones polynomial can be computed from the representation theory of $\mathcal{U}_q(\mathfrak{sl}(2))$. The Alexander polynomial has an analogous description in terms of the representation theory of ...

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92 views

### Non-graded representations over Lie superalgebra $\mathfrak{gl}(m,n)$

I have the following questions:
Let $m,n$ be positive integers. Consider representations over the general linear Lie super-algebra $\mathfrak{gl}(m,n)$. Namely, modules over the associative algebra ...

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141 views

### Orthosymplectic group, matrix representations

We have the orthosymplectic $osp(n,m|2k)$. The bosonic part is $so(n,m)\times sp(2k)$. The lie algebra generators are given in eg
http://cds.cern.ch/record/524737/files/0110257.pdf$
where the group ...

**3**

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**1**answer

88 views

### homomorphism of Lie superalgebras

In the book Shun-Jen Cheng, Weiqiang Wang Dualities and Representations of Lie Superalgebrasm. One founds the following definition(Definition 1.3):
Let $\mathfrak{g}$ and $\mathfrak{g'}$ be Lie ...

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**1**answer

326 views

### Kauffman's state model for the Alexander polynomial, via representation theory

I've been reading Oleg Viro's paper on "quantum relatives of the Alexander polynomial" (arXiv:math/0204290), which, among other more general things, derives state-sum formulas for the Alexander ...

**3**

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**1**answer

146 views

### Linear independence in (graded) Lie algebras

I asked a mixed-up version of this question earlier.
The Lie algebras I have in mind are the homotopy Lie algebras of wedges of finitely many spheres (in dimensions greater than $1$). Thus each ...

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**1**answer

169 views

### Definition of the supertrace in superalgebra representations

Let us consider a matrix superalgebra $A$ with generators satisfying $[L_a,L_b]=i L_c f^c{}_{ab}.$ The generators are matrices on which supertrace is defined bu the usual trace on the bosonic part ...

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### What are the simple Lie superalgebras of type E?

Background
Simple finite dimensional Lie superalgebras over $\Bbb C$ have been classified. There are the Cartan type superalgebras (algebras of purely odd vector fields), two strange families P(n) ...

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345 views

### Lie superalgebra in two dimensions

The standard formulation of two dimensional $N=(2,2)$ and $N=(0,2)$ supersymmetry algebras in physics is an explicit one; I am not aware of the underlying abstract Lie superalgebras. Does anyone know ...

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76 views

### Finite dimensional consistently graded Lie superalgebras of depth greater than 2

Victor Kac, in the paper
"Classification of infinite-dimensional simple linearly compact Lie superalgebras", http://www.mat.univie.ac.at/~esiprpr/esi605.pdf
writes at the beginning of section 5 ...

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251 views

### Is the SUSY Algebra isomorphic for all Kähler Manifolds?

For a Kähler manifold, the graded algebra generated by $\partial,\overline{\partial},\partial^*,\overline{\partial}^\ast$, the Lefschetz operator, and the dual Lefschetz operator, is called the ...

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308 views

### Do all finite $W$-superalgebras have 1-dimensional representations?

Premet proved the famous KW-conjecture in modular Lie algebra.
After, Premet introduced the finite $W$-algebra $U(g, e)$.
Also, Premet proposed the conjecture every algebra $U(g, e)$ admits a ...

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**1**answer

915 views

### I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt

Question: I am talking about the proof given on pages 50-52 of Pierre Deligne, Pavel Etingof, Daniel S. Freed, Lisa C. Jeffrey, David Kazhdan, John W. Morgan, David R. Morrison, and Edward Witten ...

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**2**answers

296 views

### Building Lie-like algebras from modules over semisimple Lie algebras

Here is a construction of a very broad class of "Lie-like" algebras, and I want to know more about them.
Here is the main definition: Suppose $\mathfrak{g}$ is a complex semsimple Lie algebra and ...

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**1**answer

311 views

### Is there much theory of superalgebras acting on manifolds by alternating polyvector fields?

Usual story: vector fields on $M$, with their Lie bracket, form a Lie algebra. We can consider "actions" of some other Lie algebra ${\mathfrak g}$ on $M$ by looking at Lie homomorphisms ${\mathfrak ...

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817 views

### Is there a definition of analogue Weyl group for Lie super algebra?

I heard from some people working in Lie super algebra that there was no proper definition for Weyl group of Lie super algebra. I do not know Lie super algebra at all. But When I searched on Google, I ...